In probability theory, the Vysochanskij–Petunin inequality gives a lower bound for the probability that a random variable with finite variance lies within a certain number of standard deviations of the variable's mean, or equivalently an upper bound for the probability that it lies further away. The sole restrictions on the distribution are that it be unimodal and have finite variance; here unimodal implies that it is a continuous probability distribution except at the mode, which may have a non-zero probability.
Theorem
Let
be a random variable with unimodal distribution, and
. If we define
then for any
,
![{\displaystyle {\begin{aligned}\operatorname {Pr} (|X-\alpha |\geq r)\leq {\begin{cases}{\frac {4\rho ^{2}}{9r^{2}}}&r\geq {\sqrt {8/3}}\rho \\{\frac {4\rho ^{2}}{3r^{2}}}-{\frac {1}{3}}&r\leq {\sqrt {8/3}}\rho .\\\end{cases}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f8ca4ee4891ad3d376b4c014f212446851da557)
Relation to Gauss's inequality
Taking
equal to a mode of
yields the first case of Gauss's inequality.
Tightness of Bound
Without loss of generality, assume
and
.
- If
, the left-hand side can equal one, so the bound is useless.
- If
, the bound is tight when
with probability
and is otherwise distributed uniformly in the interval
.
- If
, the bound is tight when
with probability
and is otherwise distributed uniformly in the interval
.
Specialization to mean and variance
If
has mean
and finite, non-zero variance
, then taking
and
gives that for any
![{\displaystyle \operatorname {Pr} (\left|X-\mu \right|\geq \lambda \sigma )\leq {\frac {4}{9\lambda ^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06a628c060b3575cfecf4ba95696554343d91cb3)
Proof Sketch
For a relatively elementary proof see.[1] The rough idea behind the proof is that there are two cases: one where the mode of
is close to
compared to
, in which case we can show
, and one where the mode of
is far from
compared to
, in which case we can show
. Combining these two cases gives
When
, the two cases give the same value.
Properties
The theorem refines Chebyshev's inequality by including the factor of 4/9, made possible by the condition that the distribution be unimodal.
It is common, in the construction of control charts and other statistical heuristics, to set λ = 3, corresponding to an upper probability bound of 4/81= 0.04938..., and to construct 3-sigma limits to bound nearly all (i.e. 95%) of the values of a process output. Without unimodality Chebyshev's inequality would give a looser bound of 1/9 = 0.11111....
One-sided version
An improved version of the Vysochanskij-Petunin inequality for one-sided tail bounds exists. For a unimodal random variable
with mean
and variance
, and
, the one-sided Vysochanskij-Petunin inequality[2] holds as follows:
![{\displaystyle \mathbb {P} (X-\mu \geq r)\leq {\begin{cases}{\dfrac {4}{9}}{\dfrac {\sigma ^{2}}{r^{2}+\sigma ^{2}}}&{\mbox{for }}r^{2}\geq {\dfrac {5}{3}}\sigma ^{2},\\{\dfrac {4}{3}}{\dfrac {\sigma ^{2}}{r^{2}+\sigma ^{2}}}-{\dfrac {1}{3}}&{\mbox{otherwise.}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6681f5a011be4106a0aac3948f1576a339c6284c)
The one-sided Vysochanskij-Petunin inequality, as well as the related Cantelli inequality, can for instance be relevant in the financial area, in the sense of "how bad can losses get."
Proof
The proof is very similar to that of Cantelli's inequality. For any
,
![{\displaystyle {\begin{aligned}\mathbb {P} (X-\mu \geq r)&=\mathbb {P} ((X+u)-\mu \geq r+u)\\&\leq \mathbb {P} (|(X+u)-\mu )|\geq r+u).\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17cc29658450df9084602abd9a7e9db2a5238bd3)
Then we can apply the Vysochanskij-Petunin inequality. With
, we have:
![{\displaystyle {\begin{aligned}\mathbb {P} (|(X+u)-\mu )|\geq r+u)&\leq {\begin{cases}{\frac {4}{9}}{\frac {\rho ^{2}}{(r+u)^{2}}}&r+u\geq {\sqrt {8/3}}\rho \\{\frac {4}{3}}{\frac {\rho ^{2}}{(r+u)^{2}}}-{\frac {1}{3}}&r+u\leq {\sqrt {8/3}}\rho \end{cases}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15ca0057b33a809ccbcd54dc40818057238f97ce)
As in the proof of Cantelli's inequality, it can be shown that the minimum of
over all
is achieved at
. Plugging in this value of
and simplifying yields the desired inequality.
Generalisation
Dharmadhikari and Joag-Dev [3] generalised the VP inequality to deviations from an arbitrary point and moments of order
other than
![{\displaystyle {\begin{aligned}P(|X-\alpha |\geq r)\leq \max \left\{{\frac {s\tau _{k}-r^{k}}{(s-1)r^{k}}},\left[{\frac {k}{k+1}}\right]^{k}{\frac {\tau _{k}}{r^{k}}}\right\}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b24cb0a59bb3d37ef2138aee85a2dacf3453742)
where
![{\displaystyle {\begin{aligned}\tau _{k}=E\left(|X-\alpha |^{k}\right),s>(k+1),s(s-k-1)^{k}=k^{k}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3664aa610fe975886cebc404fb7f05d1830677f9)
The standard form of the inequality can be recovered by setting
which leads to a unique value of
.
See also
- Gauss's inequality, a similar result for the distance from the mode rather than the mean
- Rule of three (statistics), a similar result for the Bernoulli distribution
References
- ^ Pukelsheim, F., 1994. The Three Sigma Rule. The American Statistician, 48(2), pp.88-91
- ^ Mercadier, Mathieu; Strobel, Frank (2021-11-16). "A one-sided Vysochanskii-Petunin inequality with financial applications" (PDF). European Journal of Operational Research. 295 (1): 374–377. doi:10.1016/j.ejor.2021.02.041. ISSN 0377-2217.
- ^ Dharmadhikari, S.W. and Joag-Dev, K., 1986. The Gauss–Tchebyshev inequality for unimodal distributions. Theory of Probability & Its Applications, 30(4), pp.867-871.
- D. F. Vysochanskij, Y. I. Petunin (1980). "Justification of the 3σ rule for unimodal distributions". Theory of Probability and Mathematical Statistics. 21: 25–36.
- Report (on cancer diagnosis) by Petunin and others stating theorem in English